Probability calculation: Bulbs are produced in $3$ locations.. 
Bulbs are produced in $3$ different locations $I,II,III$. Every hour,
  $I$ produces $300$ bulbs, $II$ produces $200$ bulbs and $III$ produces
  $500$ bulbs. $B$ is the event that represents a defective bulb that
  has been picked out. It's also known that of each $1000$ produced
  bulbs, $15$ bulbs of $I$, $6$ of $II$ and $45$ of $III$ are defective.
  Determine the probability $P(A_i \cap B)$ for the events $A_i \cap B$

So I thought it would be helpful to firstly calculate the probability to get a bulb from $I, II, III$
In total we have $1000$ bulbs produced every hour, $300$ of them in $I$, $200$ in $II$, $500$ in $III$. Thus we have probabilities 
$P(A_1)= \frac{300}{1000}= 30$%, $\text{ }$ $P(A_2)= \frac{200}{1000}= 20$%, $\text{ }$ $P(A_3)= \frac{500}{1000}= 50$%
The probability  we pick a defective bulb from these $1000$ is $\frac{15+6+45}{100}= \frac{33}{500}= 6.6$%
Now we are supposed to calculate $P(A_i \cap B)$. What does that mean? I think that notation is equal to $P(A_i)-P(B)$. Thus we have $P(A_1)-P(B) = 0.3 - 0.066= 0.234 = 23.4$% and we apply the same for $i=2, i=3$
Is it right like that? If not I hope you can show me how it's done correctly?
 A: The notation $\cap$  means intersection. 
$A_i \cap B$ is the event that both $A_i$ happens and $B$ happens.
In general $P(A_i \cap B ) \neq P(A_i) - P(B)$
If $A_i$ means a bulb is produced in location $I$.
$Pr(A_1 \cap B)$ is the probability that a bulb was produced in location $I$ and it is defective.
A: First it must be cleared what is the meaning of "..of each $1000$ produced bulbs, $15$ bulbs of I .." . Normally speaking that should mean 


*

*a) ".. out of $1000$ bulbs produced by each plant, ...", (i.e. we give the defective rate of each single plant) and not

*b)".. out of $1000$ bulbs totally produced.


Supposing a) is the correct interpretation,
then every hour I produces $300$ bulbs, $15/1000$ of which are defective, so $=4.5$ defective/h, II will produce $200 \cdot 6/1000=1.2 d/h$ and III will produce $500 \cdot 45/1000=22.5 d/h$.
Thus totally there are $D=28.2 d/h$ and the probabilities required will clearly be $4.5/D,\; 1.2/D,\; 22.5/D$.
In the interpretation b) instead
$D$ would be $66 d/h$, and the probabilities $15/D,\; 6/D,\; 45/D$.
